Question

applying sridhar acharya method solve :
1/x - 2 + 1/x - 3 + 1/x - 4 = 0​

Answer 1

GIVEN :–

$\\ \bf \implies \dfrac{1}{x - 2} + \dfrac{1}{x - 3} +\dfrac{1}{x - 4} = 0$

TO FIND :–

• Value of 'x' = ?

SOLUTION :–

$\\ \bf \implies \dfrac{1}{x - 2} + \dfrac{1}{x - 3} +\dfrac{1}{x - 4} = 0$

$\\ \bf \implies \dfrac{(x - 3) + (x - 2)}{(x - 2)(x - 3)} +\dfrac{1}{x - 4} = 0$

$\\ \bf \implies \dfrac{2x - 5}{(x - 2)(x - 3)} +\dfrac{1}{x - 4} = 0$

$\\ \bf \implies \dfrac{2x - 5}{ {x}^{2} - 5x + 6} +\dfrac{1}{x - 4} = 0$

$\\ \bf \implies \dfrac{(2x - 5)(x - 4) +{x}^{2} - 5x + 6 }{ ({x}^{2} - 5x + 6)(x - 4)}= 0$

$\\ \bf \implies \dfrac{2 {x}^{2} - 13x + 20 +{x}^{2} - 5x + 6 }{ ({x}^{2} - 5x + 6)(x - 4)}= 0$

$\\ \bf \implies \dfrac{2 {x}^{2} - 13x + 20 +{x}^{2} - 5x + 6 }{(x - 2)(x - 3)(x - 4)}= 0$

$\\ \bf \implies \dfrac{3{x}^{2} - 18x + 26}{(x - 2)(x - 3)(x - 4)}= 0$

• We should write this as –

$\\ \bf \implies 3{x}^{2} - 18x + 26= 0$

$\\ \bf \: \: \bigg[ \: \: \because \: \: x \ne \: 2,3,5 \bigg]$

• Now using sridharacharya formula –

$\\ \bf \implies x = \dfrac{ - ( - 18) \pm \sqrt{ {( - 18)}^{2} - 4(3)(26)} }{2(3)}$

$\\ \bf \implies x = \dfrac{ 18 \pm \sqrt{324-312} }{6}$

$\\ \bf \implies x = \dfrac{ 18 \pm \sqrt{12} }{6}$

$\\ \bf \implies x =3 \pm\dfrac{1}{ \sqrt{3} }$

$\\ \large\implies \pink{ \boxed{ \bf{x = \left(3 + \dfrac{1}{ \sqrt{3} } \right) , \left(3 - \dfrac{1}{ \sqrt{3}} \right)}}}$

Answer 2

Answer:

Given :-

• $\sf \dfrac{1}{x - 2} + \dfrac{1}{x - 3} + \dfrac{1}{x - 4} =\: 0$

To Find :-

• What is the value of x.

Solution :-

$\longmapsto$ $\sf \dfrac{1}{x - 2} + \dfrac{1}{x - 3} + \dfrac{1}{x - 4} =\: 0$

$\implies \sf \dfrac{(x - 3)(x - 4) + (x - 2)(x - 4) + (x - 2)(x - 3)}{(x - 2)(x - 3)(x - 4)} =\: 0$

$\implies \sf \dfrac{{x}^{2} - 4x - 3x + 12 + {x}^{2} - 4x - 2x + 8 + {x}^{2} - 3x - 2x + 6}{(x - 2)(x - 3)(x - 4)} =\: 0$

$\implies \sf \dfrac{{x}^{2} - 7x + 12 + {x}^{2} - 6x + 8 + {x}^{2} - 5x + 6}{(x - 2)(x - 3)(x - 4)} =\: 0$

$\implies \sf \dfrac{{x}^{2} + {x}^{2} + {x}^{2} - 7x - 6x - 5x + 12 + 8 + 6}{(x - 2)(x - 3)(x - 4)} =\: 0$

$\implies \sf \dfrac{{3x}^{2} - 18x + 26}{(x - 2)(x - 3)(x - 4)} =\: 0$

By doing cross multiplication we get,

$\implies \sf {3x}^{2} - 18x + 26 =\: 0$

From this quadratic equation we have to find the value of a , b and c then,

• a = 3
• b = - 18
• c = 26

Now, we have to apply Sridhar Acharya Method, we know that,

$\sf\boxed{\bold{\pink{x =\: \dfrac{- b \pm \sqrt{{b}^{2} - 4ac}}{2a}}}}$

Now, by putting the value of a , b and c we get,

$\implies \sf x =\: \dfrac{-(- 18) \pm \sqrt{{(- 18)}^{2} - 4(3)(26)}}{2(3)}$

$\implies \sf x =\: \dfrac{18 \pm \sqrt{324 - 4 \times 78}}{(2 \times 3}$

$\implies \sf x =\: \dfrac{18 \pm \sqrt{324 - 312}}{6}$

$\implies \sf x =\: \dfrac{18 \pm \sqrt{12}}{6}$

$\implies \sf x =\: 3 \pm \dfrac{1}{\sqrt{3}}$

Then,

$\implies \sf\bold{x =\: 3 + \dfrac{1}{\sqrt{3}}}$

Either,

$\implies \sf\bold{x =\: 3 - \dfrac{1}{\sqrt{3}}}$

$\therefore$ $\sf\boxed{\bold{\green{The\: value\: of\: x\: is\: 3 + \dfrac{1}{\sqrt{3}} ,\: 3 - \dfrac{1}{\sqrt{3}}.}}}$